Vectors and 3D Coordinate Geometry

1. The problem statement, all variables and given/known data
Suppose that II c R3 is a plane, and that P is a point not on II. Assume that Q is a point in II whose distance to P is minimal; in other words, the distance from P to Q is less than or equal to the distance from P to any other point in II. Show that the vector PQ is orthogonal to II.
Hint given: Define a differentiable vector function r(t) with r(t) a subset of II, and r(0) = Q. Let p be the vector with components given by the coordinates of P. Let
f(t) = |r(t)-p|2 = (r(t)-p) . (r(t)-p)
What can u say about
df(t)/dt |t=0 ?

A very simple way to do this is purely geometric. Suppose PQ is NOT perpendicular to the plane. Draw the perpendicular from P to the plane and call the point at which the perependicular crosses the plane R. Then PQR is a right triangle with right angle at R. PR is a leg of that right triangle and PQ is the hypotenuse. By the Pythagorean theorem, [itex]|PQ|= \sqrt{|PR|^2+ |QR|^2}< |PR|[/itex] contradicting the fact that |PQ| is minimal. (| | here indicates the length of the line.)

A very simple way to do this is purely geometric. Suppose PQ is NOT perpendicular to the plane. Draw the perpendicular from P to the plane and call the point at which the perependicular crosses the plane R. Then PQR is a right triangle with right angle at R. PR is a leg of that right triangle and PQ is the hypotenuse. By the Pythagorean theorem, contradicting the fact that |PQ| is minimal. (| | here indicates the length of the line.)

Oh wow, that method's a lot simpler :). Thanks HallsOfIvy.

Tiny-tim, is my solution correct now:
Let r(t) be a differentiable vector function r(t) with r(t) a subset of II, and r(0) = Q.
Let p be the vector with components given by the coordinates of P.
Let f(t) = |r(t)-p|2 = (r(t)-p) . (r(t)-p)
So, f'(t) = (r(t)-p).r'(t) + (r(t)-p).r'(t) = 2(r(t)-p).r'(t)
Since the distance from P to Q is less than or equal to the distance from P to any other point in II, f'(t) = 0 when r(t) = Q. This occurs when t=0 by definition
So f'(0) = 0
But f'(0) = 2(r(0)-p).r'(0) = 2(Q-p).r'(0)
Therefore, 2(Q-p).r'(0) = 2(PQ).r'(0) = 0
Since r'(0) lies in II, PQ is orthogonal to II

Tiny-tim, is my solution correct now:
Let r(t) be a differentiable vector function r(t) with r(t) a subset of II, and r(0) = Q.
Let p be the vector with components given by the coordinates of P.
Let f(t) = |r(t)-p|2 = (r(t)-p) . (r(t)-p)
So, f'(t) = (r(t)-p).r'(t) + (r(t)-p).r'(t) = 2(r(t)-p).r'(t)
Since the distance from P to Q is less than or equal to the distance from P to any other point in II, f'(t) = 0 when r(t) = Q. This occurs when t=0 by definition
So f'(0) = 0
But f'(0) = 2(r(0)-p).r'(0) = 2(Q-p).r'(0)
Therefore, 2(Q-p).r'(0) = 2(PQ).r'(0) = 0
Since r'(0) lies in II, PQ is orthogonal to II

Hi kehler!

Yes, that's fine, except for the last line.

All you've proved is that PQ is perpendicular to the tangent of that particular curve, r(t).

So you should add "but this applies for any curve through Q, and so PQ is perpendicular to the tangent of every curve in II through Q, ad so is perpendicular to II."

btw, HallsofIvy's proof works fine in this case, where II is a plane, but the proof above will work for any surface, to prove that pQ is perpendicular to the tangent plane at Q. ​